Matrices & Determinants
Determinant of adj — Properties of Determinants
nta_pyq_2024_apr
Grade 12

Question:

Let $\alpha\in(0,\infty)$ and $A=\begin{bmatrix}1&2&\alpha\\1&0&1\\0&1&2\end{bmatrix}$. If $\det(\text{adj}(2A-A^T)\cdot\text{adj}(A-2A^T))=2^8$, then $(\det(A))^2$ is equal to:
36
16
1
49

Step-by-Step Solution

Key Concept: $|\text{adj}(2A-A^T)\cdot\text{adj}(A-2A^T)|=|(2A-A^T)|^2\cdot|(A-2A^T)|^2=2^8$. Note $(A-2A^T)^T=A^T-2A$, so $|A-2A^T|=|A^T-2A|$, giving $|A-2A^T|^2=16$, $|A-2A^T|=\pm4$.
$\alpha=1$, $|A|=-4$, $(\det A)^2=16$.
Correct Answer: 2

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